oj.uz

Submission #8555

# Submission time Handle Problem Language Result Execution time Memory
8555 2014-09-17T04:06:05 Z a555 팩토리얼 세제곱들의 합 (YDX14_fact) C++
0 / 1
 0 ms 1676 KB 
#include <iostream>
#include <iomanip>
#include <stdlib.h>
#include <stdio.h>
#include <cstring>
#include <cstdlib>
#include <ctime>
#include <algorithm>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <map>
#include <math.h>
#include <bitset>
#include <numeric>



#pragma warning(disable:4996)

#define REP(variable, repeatnumber) for(int variable=0; variable<(repeatnumber); ++variable)
#define FOR(variable, start, end) for(int variable=(start); variable<=(end); ++variable)
#define RFOR(variable, start, end) for(int variable=(start); variable>=(end); --variable)
#define ULL unsigned long long
#define LL long long 
using namespace std; // 700B

int n, k;
int tb[6][4];

void kzro(){
	int s = 1;
	REP(i, n + 1) s += i;
	while (1){
		if (s % 10 != 0){
			printf("%d\n", s % 10);
			break;
		}
		else s /= 10;
	}
}

int main()
{
	scanf("%d%d", &n, &k);
	if (k == 0){
		kzro();
	}
	else if (n == 0 || (n == 2 && k == 3) || (n == 3 && k == 1)) 
		printf("1\n");
	else if (n == 1 || (n == 3 && k == 2)) printf("2\n");
	else if (n == 2 && k == 1) printf("4\n");
	else if (n == 3 && k == 3) printf("6\n");
	else if (n == 2 && k == 2) printf("8\n");
	else if ((n == 4 || n == 5 ) && k == 1) printf("4\n");
	else if ((n == 4 || n == 5) && k == 2) printf("8\n");
	else if ((n == 4 || n == 5) && k == 3) printf("5\n");

	return 0;
}

Subtask #1
0.0 / 1.0

# Verdict Execution time Memory Grader output
1 Correct 0 ms 1676 KB Output is correct - answer is '4'
2 Correct 0 ms 1676 KB Output is correct - answer is '2'
3 Correct 0 ms 1676 KB Output is correct - answer is '2'
4 Correct 0 ms 1676 KB Output is correct - answer is '2'
5 Correct 0 ms 1676 KB Output is correct - answer is '2'
6 Correct 0 ms 1676 KB Output is correct - answer is '4'
7 Incorrect 0 ms 1676 KB Output isn't correct - expected '6', found '8'
8 Halted 0 ms 0 KB -